Question: Let $g(x)=\sin(2x)$, for $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$. Where does $g$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-\dfrac{\pi}{4}$ (Choice B) B $x=0$ (Choice C) C $x=\dfrac{\pi}{4}$ (Choice D) D $g$ has no critical points.
Explanation: A critical point of $g$ is a point in the domain of $g$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $g$, let's find its derivative. $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[ \sin(2x) \right] \\\\ &=\cos(2x) \cdot \dfrac{d}{dx}[2x] \\\\ &=2\cos(2x) \end{aligned}$ Now let's look for $x$ -values where $g'$ is zero or undefined. In the interval $-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}$, $2\cos(2x)=0$ at $x=-\dfrac{\pi}{4}$ and $x=\dfrac{\pi}{4}$. $2\cos(2x)$ is never undefined, so $g'$ is never undefined. In conclusion, these are the $x$ -values where $g$ has critical points: $x=-\dfrac{\pi}{4}$ $x=\dfrac{\pi}{4}$